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Home , Trace class, Ultraweak topology

THE PREDUAL OF A .

A.P.M. KUPERS

Contents 1. The predual 1 2. The case B(H): -class and Hilbert-Schmidt operators 2 3. Standard forms for II1 factors 5

We will always assume that our Hilbert spaces are separable.

1. The predual Let’s fix some notation: if B is a , then B∗ is the vector space of -continuous ∗ linear functionals. This has two possible topologies: the topology (B)n and the ∗ weak-* topology (B)w. In general we will look at the latter.

Let A be a von Neumann algebra. The goal of this talk will be naturally associate to a pair ∗ (A, A ,→ B(H)) a Banach space A∗ such that A = (A∗)w, where the former is considered as having the as a subspace of B(H). We claim that A∗ can be given by the space of ultraweakly continuous functionals. We will show ∗ ∼ that this is a Banach space and that (A∗)w = A. The Banach space A∗ will be called the predual. ∗ To get a better feeling for A∗, let A be the vector space of norm continuous linear functionals φ : A → C (with induced norm). Because norm convergence implies ultraweak convergence, it ∗ follows that A∗ is a subspace of A .

We will reduce the proof to the case that A = B(H) using the following results. Recall that Hahn-Banach says that if one has a linear functional φ on a subspace of a vector space dominated by a sublinear functional N (think of it as a norm), then one can extend φ to the entire space in a way that it is still dominated by N . Proposition 1.1. If x ∈ B(H) satisfies the φ(x) = 0 for each ultraweakly continuous φ with φ(A) = 0, then x ∈ A. Proof. This works for V any ultraweakly closed subspace of B(H). If x∈ / V , then apply Hahn- Banach or one of its corollaries (e.g. Conway IV.3.15) to the linear functional φ defined by φ(x) = 1 and φ(v) = 0 for all v ∈ V . This is continuous with respect to the restriction of the ultraweak topology on V + Cx, hence extends to a ultraweakly continuous functional φ on B(H) with the property that φ(x) = 1.  We prove the reduction to the case B(H) in the next section. However, we show that the embedding in B(H) is not essential for the theorem, by given an intrinsic definition of the predual. ∼ ∗ ∼ ⊥ ∗ ⊥ Theorem 1.2. Suppose that B(H)uw = (B)w. Then Auw = (B/A )w and furthermore B/A can be identified as a vector space with the subspace A∗ of ultraweakly continuous linear functionals on A.

Date: March 20, 2011. 1 2 A.P.M. KUPERS

⊥ ∗ Proof. We first identify Auw with (B/A )w. By definition there is an continuous embedding ∗ ∗ Auw → B(H)uw = Bw. Because A is ultraweak closed, it is weak-* closed in B(H) = (B)w. This implies that there is a homeomorphism between the subspace of the norm-continuous functionals ⊥ ∗ vanishing on A with the subspace topology of the weak-* one, and (B/A )w. This can be seen ⊥ ∗ as follows: the embedding A → (B/A )w is clearly continuous, open and injective, and surjective using the previous proposition. Next we identify the vectorspace (B/A⊥)∗ with the space of ultraweakly continuous functionals ⊥ on A. This means that we must identify the B/A with the weak-* continuous linear functionals A∗ ⊥ ∗∗ ∗ ⊥ in the double dual (B/A ) = (Auw) . See section V.1 of Conway: clearly the elements of B/A ∗ are weak-* continuous considered as elements of (Auw) . For the converse, one notes that a weak-* continous functional is bounded in absolute value by a finite sum of iv = b 7→ |b(v)| for ⊥ v ∈ B/A and inductively use this to write the functional as a linear combination of the iv’s.  2. The case B(H): trace-class and Hilbert-Schmidt operators In this section we will see that the predual of B(H) is given by the Banach space T (H) of traceclass operators with the trace-norm | − |1.

Definition 2.1. Let a ∈ B(H) and let (ξi) be an . Then a is said to be of if the following sum is finite X |a|1 = h|a|ξi, ξii i The set of trace-class operators is denoted by T (H). To show that this is well-defined, one proves that the sum is independent of the choice of orthonormal basis.

Lemma 2.2. |a|1 is independent of the choice of orthonormal basis. Proof. Note that |a| = p|a|p|a| and write X X p p h|a|ξi, ξii = h |a|ξi, |a|ξii i i X X p 2 = |h |a|ξi, ηii| i j X X p 2 = |h |a|ηi, ξii| j i X = h|a|ηj, ηji j where we were allowed to interchange the sum because all terms are positive.  We now discuss several useful properties of the trace-class operators.

Proposition 2.3. T (H) is an ideal which is closed under ∗. The map | − |1 is a complete norm on T (H) satisfying ||a|| ≤ |a|1, making it into a Banach space. The finite rank operators are dense and hence T (H) is in fact a sub-ideal of the compact operators K(H).

Proof. To show that it is an ideal in B(H), use that |xa|1 ≤ ||x|||a|1 for x ∈ B(H) and similarly ∗ |ax|1 ≤ ||x|||a|1. That it is closed under ∗ is clear, as |a| = |a |. p 2 To show that ||a|| ≤ |a|1 one remarks that ||a|| ≤ || |a||| ≤ |a|1. The first inequality follows √ 2 from the . For the second inequality, note that || a|| = sup||ξ||=1h|a|ξ, ξi and hence we pick any ξ of norm 1 for which this supremum is almost obtained as the first vector of p 2 an orthonormal basis to bound |a|1 ≥ || |a||| −  for all  > 0. Taking  to zero then proves the statement. To show that | − |1 is a complete norm, one notes that if an is | − |1-Cauchy, it is || − ||-Cauchy to get a candidate a. We need to show that a is trace-class and |a−an|1 → 0. However, the second statement implies the first since |a|1 ≤ |an|1 + |a − an|1. So let’s prove the second: if ||a − an|| → 0, THE PREDUAL OF A VON NEUMANN ALGEBRA. 3 then in particular |am − an|ξi → |a − an|ξi for all vectors ξi of length 1. This implies that if P |am − an|1 = h|am − an|ξi, ξii ≤  for m, n large enough, then |a − an|1 ≤  as well. Finally, note that any a ∈ T (H) is the | − |1-limit of apn, where pn is the projection on the first n basis vectors ξi for 1 ≤ i ≤ n. All apn are of finite rank, hence these are dense. Because | − |1 dominates || − ||, T (H) lies in the norm-closure of the finite-rank operators, which are the compact operators.  Next we define the trace. Lemma 2.4. If a is a trace-class operator, then the following sum converges absolutely X Tr(a) = haξi, ξii i

It satisfies Tr(a) ≤ |a|1 and Tr(xa) ≤ ||x||Tr(a) for each x ∈ B(H). Proof. It suffices to prove the second statement. We will only prove the first part, as the second is then easy. To prove this, write a = up|a|p|a| using the and consider the following: p p ∗ 2 0 ≤ ||( |a| − λ |a|u )ξi|| p 2  p p ∗  2 p ∗ 2 = || |a|ξi|| − 2Re λh |a|ξi, |a|u ξii + |λ| || |a|u ξi|| 2 ∗ ∗ = h|a|ξi, ξii − 2Re (λhaξi, ξii) + |λ| h|a|u ξi, u ξii iφ iφ  Now write λ = e , where the phase is chosen such that Re e haξi, ξii = |haξi, ξii|. Then we get: ∗ ∗ 2|haξi, ξii| ≤ h|a|ξi, ξii + h|a|u ξi, u ξii

Now note that the sequences (ξi) and (uξi) both orthonormal bases. Thus summing over i we obtain Tr(a) ≤ |a|1  We will now prove that there is an of Banach spaces between T (H)∗ with the operator norm topology and B(H) with the norm topology. An important role will be played by the map φ : B(H) → T (H)∗ given by

φx(a) = Tr(xa) ∗ ∼ Theorem 2.5. We have that (T (H))n = B(H)n.

Proof. By the previous proposition, we know that each φx is norm continuous. Thus, only the following statement is left: if λ : T (H) → C is a linear functional which is bounded with respect to | − |1 then there is a x ∈ B(H) such that λ(a) = φx(a) and ||λ|| = ||x||. Note that we have a on B(H):

hξ, ηiλ = λ(v 7→ hv, ξiη)

By Riesz representability, hξ, ηiλ = hxξ, ηi for some x ∈ B(H) with

||x|| = sup |hξ, ηiλ| = sup |λ(v 7→ hv, ξiη)| = ||λ|| ||ξ||=||η||=1 ||ξ||=||η||=1 where only the last equality is not obvious. Clearly the left hand side is less than or equal to the right hand side, because we should evaluate on more trace-class operators to get the usual definition. However, because |λ(−)| is continuous it suffices to check it on the dense subspace of finite rank operators. Linearity then implies we can restrict to trace-class operators of the form v 7→ hv, ξiη if we want to show that the left hand side is greater than or equal to the right hand side. To show that λ(h) = φx(h), it suffices to note that both are continuous with respect to | − |1 and coincide on finite rank operators, thus must be equal.  4 A.P.M. KUPERS

Of course, we are more interested in the weak-* and ultraweak topologies. We claim that the same statement holds with these. To prove this we will need to introduce the Hilbert-Schmidt operators. The idea is that every trace-class operator splits as a product of Hilbert-Schmidt operators and Hilbert-Schmidt operators sends an orthonormal basis to a `2-convergent sequence of vectors, hence allowing us to make the connection with the ultraweak topology. One should think of a Hilbert-Schmidt operator as a square root of a trace-class operator. Definition 2.6. A b ∈ B(H) is said to be Hilbert-Schmidt if b∗b is trace-class. We let HS(H) denote the set of Hilbert-Schmidt operators. We will need the following lemma’s, which are elementary, but very useful for calculations. Lemma 2.7. Every trace-class operator is Hilbert-Schmidt and in fact every trace-class operator 2 is a product of two Hilbert-Schmidt operators. Every ` -convergent sequence (ηi) of vectors is the image of a basis (ξi) under a Hilbert-Schmidt operator. Proof. If a is a trace-class operator, then a∗a is trace-class as well, because T (H) is an ideal. Let a = u|a| be the polar decomposition of a trace-class operator. Then a can be written as the product of trace-class operators up|a| and p|a|. P For the second assertion, note that ih−, ξiiηi is a Hilbert-Schmidt operator.  Lemma 2.8. For b, c ∈ HS(H), bc ∈ T (H) and Tr(bc) = Tr(cb). Proof. For the first statement, we note that in our proof that the trace is well-defined, we have P 2 2 shown that Tr(bc) ≤ i ||bξi|| + ||cξi|| . It is easy to check that b ∈ HS(H) is equivalent to P 2 i ||bξi|| < ∞. pP 2 The second statement one proves that | − |2 given by |b|2 = i ||bξi|| is a complete norm for the Hilbert-Schmidt operators such that the finite-rank operators are dense and Tr is continuous in both components. The equation then follows by continuity from the corresponding equation for finite-rank operators, where it is the usual invariance under cyclic permutation of the trace of a matrix.  We can now prove the result about the ultraweak topology on B(H) and the weak-* topology on T (H). ∗ ∼ Theorem 2.9. We have that (T (H))w = B(H)uw. Proof. The only thing that is left to do is prove that ultraweak topology on B(H) comes from the seminorms x 7→ |Tr(xa)| for a ∈ T (H). The ultraweak topology is obtained by looking at 2 seminorms | − |ξ,η for sequences (ξi), (ηi) ∈ ` (H) given by ∞ X | − |ξ,η = |hxξi, ηii| i=1 P∞ To relate this to the |x|h = |Tr(xh)| = | i=1hxhζi, ζii| with (ζi)i an orthonormal ∗ ∗ basis. We note that h = h1h2 for h1, h2 Hilbert-Schmidt operators. Then we have that Tr(xh1h2) = ∗ Tr(h2xh1). Send h2 to the other side of the inner product and note that both elements (ξi), (ηi) ∈ `2(H) can be obtained by applying a Hilbert-Schmidt operator to the fixed set of orthonormal vectors (ζi)i. By including appropriate phases, we can also get the absolute values.  Finally, we show that T (H) can be identified with the ultraweakly continuous functionals on ∗ B(H). To do, we take the of φ. This is a map ψ : T (H) → B(H) given by ψa(x) = Tr(xa), where the latter has the operator norm topology. We will show that the image of this map is the space of functionals which are ultraweakly continuous in addition to being norm continuous.

Theorem 2.10. Each ψa is ultraweakly continuous and the map a 7→ ψa is continuous. To be precise, ||ψa|| = |a|1. Conversely, if ω is an ultraweakly continuous linear functional on B(H), then there is a trace- class operator a such that ω = ψa and ||ω|| = |a|1. THE PREDUAL OF A VON NEUMANN ALGEBRA. 5

Proof. For the statement about ultraweak continuity, use statement that a functional ω is ultra- weakly continuous if and only if it is of the form X ω(x) = hxξi, ηi i 2 for (ξi), (ηi) ` -convergent sequences of vectors. By the same reasoning as in the previous theorem, this exactly means that ω of the form Tr(a). To prove that ||ψa|| = |a|1, we use that because |a| is compact and positive, there exists a orthonormal basis ξi of H consisting of eigenvectors for |a| with eigenvalues λi ≥ 0. A simple ∗ estimate shows using this basis shows that ψa(x) ≤ ||x|||a|1 and entering x = u for the polar decomposition a = u|a| shows equality. For the converse statement one uses similar reasoning as in the proposition. If ω is ultraweakly ∗ continuous then it can be written as ω(x) = Tr(a2xa1) for a1, a2 ∈ HS(H). Use the cyclic property of the trace to show that ω(x) = ψa1a2 (x). 

The conclusion is that T (H) is the norm topology from | − |1 is isometrically isomorphic to the subspace of B(H)∗ of the ultraweakly continuous functionals with the subspace topology from the operator norm topology. Furthermore, the weak-* topology on B(H) is exactly the ultraweak topology.

3. Standard forms for II1 factors

A is a type II1 factor if it is an infinite dimensional factor and has a ultraweakly continuous trace tr, i.e. tr(ab) = tr(ba) and tr(a∗a) ≤ 0. There is a unique normalized trace: tr(1) = 1. We want to apply the GNS-construction to this. To do this, one lets L2(A, tr) be the completion of the vector space given by A/{a ∈ A|tr(a∗a) = 0} with inner product induced by a, b 7→ tr(b∗a). The action of A on L2(A, tr) induced by multiplication in the algebra turns out to be continuous and is called the standard form of A. To link with the previous section, we note that there exist spaces Lp(A, tr) for p ∈ [1, ∞] with p 1/p norms induced by tr(| − | ) . If p = ∞, one obtain A, for p = 1, one gets the predual A∗. So, 2 one should think of L (A, tr) as the Hilbert-Schmidt operators associated to the predual A∗.